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2 years ago

How many kilograms in 1 1/2 pounds?

Mathematics

2 answers:

blondinia [14]2 years ago

6 0

Answer:

67.5 kgs.

Step-by-step explanation:

1 pound = 0.45 kg

1 1/2 pounds = 0.45 + 1/2 * 45

= 67.5 kgs.

gtnhenbr [62]2 years ago

4 0

0.68 kilograms are in 1 1/2 pounds

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Caleb is preheating his oven before using it to bake. The initial temperature of the oven is 70° and the temperature will increa

gavmur [86]

Answer:

190 Degrees Celsius

Step-by-step explanation:

so first you multiply the

Rate per minute X Minutes

15 X 8 = 120

120 this is the temperature increase so you add this increase to the initial temperature which was 70 degrees Celsius

temperature Increase + Initial temperature

120 + 70 = 190 degrees Celsius

6 0

3 years ago

When this open-ended cylinder is opened out, it forms a rectangle with a width of 25 cm. What is the area of the rectangle?.

nalin [4]

Answer:

Just want the points

Step-by-step explanation:

4 0

3 years ago

Compute using long division: 1,234÷68

Kruka [31]

Answer:

Quotient = 18

Remainder = 10

Step-by-step explanation:

1234/68

=> 68 x 1 = 68

=> 123 - 68 = 55

=> Take the 4 down

=> 554/68

=> 68 x 8 = 544

=> 554 - 544 = 10

So, the quotient = 18.

Remainder = 10

4 0

3 years ago

A company estimates that it will sell ????(x) units of a product after spending x thousand dollars on advertising, as given by ?

zepelin [54]

Answer:

$(11\frac{1}{3},20)$

Step-by-step explanation:

If the Number of Sales of x units, N(x) is defined by the function:

$N(x)=-5x^3+235x^2-3400x+20000,10\leq x\leq 40$

$N'(x)=-15x^2+470x-3400\\When -15x^2+470x-3400=0\\x=20, x=11\frac{1}{3}$

Next, we text the critical points and the end points of the interval to see where the derivative is increasing.

$N'(10)=-200\\N'(11\frac{1}{3})=0\\N'(19)=115\\N'(20)=0\\N'(40)=-8600$

Thus, the rate of change of sales $N'(x)$ is increasing in the interval $(11\frac{1}{3},20)$ on 10≤x≤40.

7 0

3 years ago

Using the function attached.<br> Find (f+G)(x)

sp2606 [1]

We have been given two functions $f(x)=-8x^3+5x^2+7x+4$ and $g(x)=8x^3-2x+3$ . We are asked to find $(f+g)(x)$ .

We will use composite function property $(f+g)(x)=f(x)+g(x)$ to solve our given problem.

$(f+g)(x)=-8x^3+5x^2+7x+4+8x^3-2x+3$

Now we will combine like terms as:

$(f+g)(x)=-8x^3+8x^3+5x^2+7x-2x+4+3$

$(f+g)(x)=5x^2+5x+7$

Therefore, the value of $(f+g)(x)$ would be $5x^2+5x+7$ .

8 0

3 years ago